Example 1 (Asymptotes)

Let

$f(x)=\frac{3x^2-2x+2}{x^2-x-2}$

What is the domain of $f$? What are the roots and $y$-intercepts of $f$? Find the horizontal and vertical asymptotes of $f$.

Solution The domain of $f$ is the set on which $f$ is defined. As a quotient of continuous functions, $f$ is defined when the denominator is not zero:

$x^2-x-2\neq0,$

$\Rightarrow (x-2)(x+1)\neq 0.$

That is the domain of $f$ is $\mathbb{R}\backslash\{-1,2\}$ (all the real numbers except -1 and 2. Note now that

$f(x)=\frac{3x^2-2x+2}{(x-2)(x+1)}$).

The roots of a quotient of continuous functions occur at points in the domain when the numerator is zero:

$3x^2-2x+2=0,$

$\Rightarrow x_{\pm}=\frac{2\pm \sqrt{4-4(3)(2)}}{6}=\frac{2\pm\sqrt{-20}}{6}\not\in\mathbb{R}.$

That is $f$ has no (real) roots ($\sqrt{-20}$ is not a real number).

The $y$-intercept of $f$ is the point where $f$ cuts the $y$-axis (i.e. when $x=0$):

$f(0)=\frac{3(0)^2-2(0)+2}{(0)^2-(0)-2}=\frac{2}{-2}=-1.$

The horizontal asymptotes of $f$ are the behaviours of $f$ as $x\rightarrow \pm\infty$:

$\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow \infty}\frac{3x^2-2x+2}{x^2-x-2}\approx \frac{3x^2}{x^2}=3$.

Hence $h(x)=3$ is the horizontal asymptote.

The vertical asymptotes of $f$ are the finite $x$ where $f$ gets very big (has a singularity). Following our procedure:

Step 1: A quotient of continuous functions can only have a vertical asymptote when the denominator is zero; i.e. $x=-1,2$ (it is necessary but not sufficient that the denominator is zero – hence Step 2).

Step 2: Evaluate the limits

$\lim_{x\rightarrow-1}f(x)$ and $\lim_{x\rightarrow 2}f(x)$

$\lim_{x\rightarrow -1}\frac{3x^2-2x+2}{x-2}.\frac{1}{x+1}=\frac{3(-1)^2-2(-1)+2}{-1-2}.\lim_{x\rightarrow -1}\frac{1}{x+1}=\frac{7}{-3}\times\pm\infty=\pm\infty.$

$\lim_{x\rightarrow 2}\frac{3x^2-2x+2}{x+1}.\frac{1}{x-2}=\frac{3(2)^2-2(2)+2}{2+1}.\lim_{x\rightarrow 2}\frac{1}{x-2}=\frac{10}{3}\times\pm\infty=\pm\infty.$

Step 3: As these limits evaluate to infinity there are vertical asymptotes at $x=-1$ and $x=2$.

Comment To carry out the full analysis of this function to sketch it is quite messy – we need a computer to find the split points of $f'$ and $f''$.

Example 2 (all elements considered)

Let

$f(x)=\frac{x^2+6x-4}{3x-2}$

Find the domain of $f$. Find the roots and $y$-intersections of $f$. Find the horizontal and vertical asymptotes of $f$. Find the intervals where $f$ is increasing/ decreasing. Find the locations of the local extrema (local max/min). Find where the function is concave up/ concave down. Hence sketch the graph.

Solution The domain of $f$ is where

$3x-2\neq 0\Leftrightarrow 3x\neq 2$

$\Leftrightarrow x\neq\frac{2}{3}$

Domain of $f$ is $\mathbb{R}\backslash\{2/3\}$.

The roots of $f$ are where

$x^2+6x-4=0$

$x_{\pm}=\frac{-6\pm \sqrt{36-4(-4)}}{2}=\frac{-6\pm\sqrt{4(13)}}{2}=-3\pm\sqrt{13}$.

The $y$ -intercept, $f(0)=\frac{-4}{-2}=2$.

The horizontal asymptotes

$\lim_{x\rightarrow\infty}\frac{x^2+6x-4}{3x-2}\approx \frac{x^2}{3x}=\frac{1}{3}x$.

Hence the horizontal asymptote $h(x)=x/3$(Which is a line of slope $1/3$ and $y$-intercept $0$. (Cf. $y=mx+c$)).

The vertical asymptotes:

Step 1: $x=2/3$.

Step 2:

$\lim_{x\rightarrow 2/3}f(x)= ((2/3)^2+6(2/3)-4).\lim_{x\rightarrow 2/3}\frac{1}{3x-2}=\frac{4}{9}\times\pm\infty=\pm\infty$

Step 3: Hence $x=2/3$ is a vertical asymptote.

To find where $f$ is increasing/ decreasing, we have to find the split points of $f'$ (the only place $f'$ can change sign; i.e. the only place where $f$ can go from increasing to decreasing or vice versa . Remember $f'>0\Rightarrow$ increasing; $f'<0\Rightarrow$ decreasing. The split points are where $f'=0$ or undefined). By the quotient rule, $f'$ is undefined when the denominator is zero; i.e. $x=2/3$. By the quotient rule:

$f'(x)=\frac{(3x-2)(2x+6)-(x^2+6x-4)(3)}{(3x-2)^2}$

$=\frac{6x^2+18x-4x-12-3x^2-18x+12}{(3x-2)^2}=\frac{3x^2-4x}{(3x-2)^2}=\frac{x(3x-4)}{(3x-2)^2}$

The other split points are where $f'=0$, and in this case this occurs when $x(3x-4)=0$; i.e. $x=0,4/3$. Now we test in between the split points to see if $f'$ is positive/negative ($f$ increasing/decreasing). The split points are $0,2/3,4/3$ so the intervals to test are:

$(-\infty,0),(0,2/3),(2/3,4/3),(4/3,\infty)$

Choose test points $-1,1/2,1,2$:

$f'(-1)=\frac{(-1)(3(-1)-4)}{(3(-1)-2)^2}=\frac{(-)(-)}{(+)}>0$

$f'(1/2)=\frac{(+)(-)}{(+)}<0$

$f'(1)=\frac{(+)(-)}{(+)}<0$

$f'(2)=\frac{(+)(+)}{(+)}>0$

Hence $f'>0$ on $(-\infty,0)\cup(4/3,\infty)$ and $f'<0$ on $(0,2/3)\cup(2/3,4/3)$; i.e. increasing on $(-\infty,0)\cup(4/3,\infty)$ and decreasing on $(0,2/3)\cup(2/3,4/3)$.

To find out the locations of the local extrema, we do the following schematic of $f$ (If $f$ is increasing on an interval draw it increasing, if $f$ is decreasing on an interval draw it decreasing. Beware of singularities! At these points $f$ is not continuous (like it is at $0$ and $4/3$)).:

(please ignore the vertical line joining them up) It is clear from this schematic that there is a local maxima at $0$ and a local minima at $4/3$. These have coordinate $(0,2)$ and $(4/3,f(4/3))=(4/3,26/9)$.

To find where $f$ is concave up/ concave down we find the split points of $f''$. By the quotient rule, as

$f'(x)=\frac{x(3x-4)}{(3x-2)^2}$,

$f''$ is undefined at $x=2/3$. By the quotient rule:

$f''(x)=\frac{(3x-2)^2(6x-4)-x(3x-4)2(3x-2)(3)}{(3x-2)^4}$

Because we have already considered the case when $x=2/3$, we may cancel the $3x-2$ (i.e. dividing by $3x-2$ is not troublesome as we have already identified $x=2/3$ as a split point and hence we will not be dividing by zero):

$f''(x)=\frac{(3x-2)(6x-4)-6x(3x-4)}{(3x-2)^3}$

$f''(x)=\frac{18x^2-12x-12x+8-18x^2+24x}{(3x-2)^3}=\frac{8}{(3x-2)^3}\neq 0,\,\forall\,x\in\mathbb{R}$

Hence the only split point of $f''$ is $x=2/3$. Now we test in between the split points to see if $f''$ is positive/negative ($f$ concave up/ down). The split point is $2/3$ so the intervals to test are:

$(-\infty,2/3),(2/3,\infty)$

Choose test points $0,1$:

$f''(0)=\frac{8}{(3(0)-2)^3}=\frac{(+)}{(-)}<0$

$f''(1)=\frac{(+)}{(+)}>0$

Hence $f''>0$ on $(2/3,\infty)$ and $f''<0$ on $(-\infty,2/3)$; i.e. concave up on on $(2/3,\infty)$ and concave down on  $(-\infty,2/3)$.

To sketch the graph take all these points into consideration (click for a sharper image):